Editing Row and column spaces (section)

===Basis===
The row space is not affected by [[elementary row operations]].  This makes it possible to use [[row reduction]] to find a [[basis (linear algebra)|basis]] for the row space.

For example, consider the matrix
:<math>A = \begin{bmatrix} 1 & 3 & 2 \\ 2 & 7 & 4 \\ 1 & 5 & 2\end{bmatrix}.</math>
The rows of this matrix span the row space, but they may not be [[linearly independent]], in which case the rows will not be a basis.  To find a basis, we reduce {{mvar|A}} to [[row echelon form]]:

{{math|'''r'''<sub>1</sub>}}, {{math|'''r'''<sub>2</sub>}}, {{math|'''r'''<sub>3</sub>}} represents the rows.
:<math>
\begin{align}
\begin{bmatrix} 1 & 3 & 2 \\ 2 & 7 & 4 \\ 1 & 5 & 2\end{bmatrix}
&\xrightarrow{\mathbf{r}_2-2\mathbf{r}_1 \to \mathbf{r}_2}
\begin{bmatrix} 1 & 3 & 2 \\ 0 & 1 & 0 \\ 1 & 5 & 2\end{bmatrix}
\xrightarrow{\mathbf{r}_3-\,\,\mathbf{r}_1 \to \mathbf{r}_3}
\begin{bmatrix} 1 & 3 & 2 \\ 0 & 1 & 0 \\ 0 & 2 & 0\end{bmatrix} \\
&\xrightarrow{\mathbf{r}_3-2\mathbf{r}_2 \to \mathbf{r}_3}
\begin{bmatrix} 1 & 3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}
\xrightarrow{\mathbf{r}_1-3\mathbf{r}_2 \to \mathbf{r}_1}
\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}.
\end{align}
</math>
Once the matrix is in echelon form, the nonzero rows are a basis for the row space.  In this case, the basis is {{math|{{mset| [1, 3, 2], [2, 7, 4] }}}}. Another possible basis {{math|{{mset| [1, 0, 2], [0, 1, 0] }}}} comes from a further reduction.<ref name="example">The example is valid over the [[real number]]s, the [[rational number]]s, and other [[number field]]s. It is not necessarily correct over fields and rings with non-zero [[characteristic (algebra)|characteristic]].</ref>

This algorithm can be used in general to find a basis for the span of a set of vectors.  If the matrix is further simplified to [[reduced row echelon form]], then the resulting basis is uniquely determined by the row space.

It is sometimes convenient to find a basis for the row space from among the rows of the original matrix instead (for example, this result is useful in giving an elementary proof that the [[Rank (linear algebra)#Alternative definitions|determinantal rank]] of a matrix is equal to its rank). Since row operations can affect linear dependence relations of the row vectors, such a basis is instead found indirectly using the fact that the column space of {{math|''A''<sup>T</sup>}} is equal to the row space of {{mvar|A}}. Using the example matrix {{mvar|A}} above, find {{math|''A''<sup>T</sup>}} and reduce it to row echelon form:

:<math>
A^{\mathrm{T}} = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 7 & 5 \\ 2 & 4 & 2\end{bmatrix} \sim 
\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0\end{bmatrix}. 
</math>

The pivots indicate that the first two columns of {{math|''A''<sup>T</sup>}} form a basis of the column space of {{math|''A''<sup>T</sup>}}. Therefore, the first two rows of {{mvar|A}} (before any row reductions) also form a basis of the row space of {{mvar|A}}.